Mechanics of Solids I

Columns

Stability of Structures

o In the design of columns, cross-sectional area is selected such that  allowable stress is not exceeded  

P A

 

all

 deformation falls within specifications  

PL AE

 

spec

o After these design calculations, one may discover that the column is unstable under loading and that it suddenly becomes sharply curved or buckles.

Euler’s Formula for Pin-Ended Beams

o Consider an axially loaded beam. After a small perturbation, the system reaches an equilibrium configuration such that

dx

2 

M EI

 

dx

2   0 o Similar to D.E. for simple harmonic motion (except for variable x instead of t) o sin cos

px

Euler’s Formula for Pin-Ended Beams

o Apply boundary conditions and thus

p



n L

 o Smallest P (critical load) is obtained for

n = 1

and

I = I min

P cr

 

L

2 2

EI

o Solution with assumed configuration can only be obtained if  

cr P A

 

L

2 2

EI

 

cr

  2   2

L r E

  2

Euler’s Formula for Pin-Ended Beams

o The value of stress corresponding to the critical load,  

cr

 

cr

 

L

2 2

EI P A

  2

L r E

  2 

cr

 

P cr A critical stress L r r

 slenderness ratio  radius of gyration o Preceding analysis is limited to centric loadings.

Extension of Euler’s Formula

o A column with one fixed and one free end, will behave as the upper-half of a pin-connected column.

o The critical loading is calculated from Euler’s formula,

P cr



cr L e

   

L

2 2

e EI

 2

E

 2  2

L

 equivalent length

Extension of Euler’s Formula

Problem 10.1

L = 0.5 m E = 70 GPa P = 20 kN FS = 2.5

o An aluminum column of length L and rectangular cross section has a fixed end at B and supports a centric load at A. Two smooth and rounded fixed plates restrain end A from moving in one of the vertical planes of symmetry but allow it to move in the other plane.

a) Determine the ratio a/b of the two sides of the cross section corresponding to the most efficient design against buckling.

b) Design the most efficient cross section for the column.

Problem 10.1

SOLUTION: The most efficient design occurs when the resistance to buckling is equal in both planes of symmetry. This occurs when the slenderness ratios are equal.

• Buckling in xy plane:

r z

2 

I z A

 1 12

ba ab

3 

a

2 12

L r z



a

0.7

L

12 • Buckling in xz plane:

r

2

y



I y A

 1 12

ab ab

3

L r y



b

2

L

/ 12 

b

12 2

r z



a

12

r y



b

12 • Most efficient design:

L r z a

0.7

L

12

a b

  

L r y

2

L b

/ 12 0.7

2

a b

 0.35

Problem 10.1

• Design: L = 0.5 m E = 70 GPa P = 20 kN FS = 2.5

a/b = 0.35